Question: When Ryan is serving at a restaurant, there is a $0.75$ probability that each party will order drinks with their meal. During one hour, Ryan served $6$ parties. Assuming that each party is equally likely to order drinks, what is the probability that at least one party will not order drinks? Round your answer to the nearest hundredth. $P(\text{at least one without drinks})=$
Strategy In this situation it is much easier to calculate the probability of the event we are looking for (at least one party that doesn't order any drinks) by calculating the probability of its complement (every party orders drinks), and subtracting from $1$. In other words, we can use this strategy: $P(\text{at least one without drinks})=1-P(\text{all }6\text{ with drinks})$ Calculations $\begin{aligned} &\phantom{=}P(\text{at least one without drinks}) \\\\ &=1-P(\text{all }6\text{ with drinks}) \\ \\ &=1-(0.75)^{6} \\ \\ &\approx 1-0.178 \\ \\ &\approx 0.822\end{aligned}$ Answer $P(\text{at least one without drinks}) \approx 0.82$